Summation of series with terms $U_n=\frac{1}{n^2-n+1} -\frac{1}{n^2+n+1}$

Question

Given that $U_n=\dfrac{1}{n^2-n+1} -\dfrac{1}{n^2+n+1}$, find $S_N$= $\sum_{n=N+1}^{2N}U_n$ in terms of $N$. Find a number $M$ such that $S_n<10^{-20}$ for all $N>M$.

I was able to calculate the sum as $\dfrac{1}{N^2+N+1}-\dfrac{1}{4N^2+2N+1}$ using the method of differences. I am having trouble doing the second part. After thinking a while I think that the correct approach is to remove some of the terms while preserving the inequality and then solve for $N$. Two such attempts were $\dfrac{1}{N^2+N+1}-\dfrac{1}{N^2}$ and $\dfrac{1}{N^2+N+1}-\dfrac{1}{N}$ but I don't know how to proceed. Is this the correct way? Any help, including hints, would be appreciated

Answer 1

You're overthinking it. Just use some crude bounds: $$\begin{align*} S_N=\frac{1}{N^2+N+1}-\frac{1}{4N^2+2N+1}&< \frac{1}{N^2+N+1}\\ &< \frac{1}{N^2}. \end{align*}$$ Therefore $M=10^{10}$ works, by which we mean that for any $N>10^{10}$, $S_N<10^{-20}$: $$ S_N<\frac{1}{N^2}<\frac{1}{10^{20}} $$

Answer 2

$$\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}=\\\frac{1}{n(n-1)+1}-\frac{1}{n(n+1)+1}=\\f(n)-f(n+1)$$ $$\sum_{n=N+1}^{2N}(f(n)-f(n+1))=f(N+1)-f(2N)$$